\( \int\left(1-x^2\right)^{-\frac{1}{2}} d x =\int \frac{1}{\left(1-x^2\right)^{1 / 2}} d x \)
\(=\int \frac{1}{\sqrt{1-x^2}} d x=\sin ^{-1} x+c\)

\(\int { \left( 1+{ x }^{ 2 } \right) ^{ -1 } } dx=\int { \frac { 1 }{ 1+{ x }^{ 2 } } } dx\)
= tan^-1 x + c

\(\int { { e }^{ x } } dx\) = e^x + c

y = tan 3x
Take u = 3x ⇒ \(\frac{d u}{d x}=3\)
\(y=\tan u\)
\( \frac{d y}{d x} =\frac{d y}{d u} \times \frac{d u}{d x}=\sec ^2 u(3)=\sec ^2(3 x) \cdot 3 \)
\(=3 \sec ^2(3 x)\)

Given y = (x² + 4x + 6)⁵
Let u = x² + 4x + 6
y = u⁵
\(\frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}\)
\(=5 u^4 \cdot(2 x+4)\)
\(=5\left(x^2+4 x+6\right)^4(2 x+4)\)

\({dy\over dx}=3x^2+10x+3.\)

\(f(x)=-x^2+2\)
\(f(x+h)=-(x+h)^2+2=-x^2-h^2-2 x h+2\)
\(f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{-x^2-h^2-2 x h+2+x^2-2}{h}\)
\(=\lim _{h \rightarrow 0} \frac{+h(-h-2 x)}{h}\)
= -0 - 2x
\(f^{\prime}(x)=-2 x\)

\(y =\sin ^3 x+\cos ^3 x \)
\(u =\sin x \quad v=\cos x\)
\(\frac{d u}{d x} =\cos x \quad \frac{d v}{d x}=-\sin x\)
\(y =u^3+v^3\)
\(\frac{d y}{d x} =3 u^2 \frac{d u}{d x}+3 v^2 \frac{d v}{d x} \)
\(=3 \sin ^2 x(\cos x)+3 \cos ^2 x(-\sin x)\)
\(=3 \sin x \cos x[\sin x-\cos x]\)

\(y=\sqrt{1+2 \tan x}\)
\(u =1+2 \tan x \)
\(\frac{d u}{d x} =2 \sec ^2 \cdot x \)
\(y =\sqrt{u}=u^{1 / 2}\)
\(\frac{d y}{d x} =\frac{d y}{d u} \cdot \frac{d u}{d x}=1 / 2 u^{1 / 2-1}\left(2 \sec ^2 x\right) \)
\(=1 / 2 u^{-1 / 2}\left(2 \sec ^2 x\right)=\frac{1}{2 \sqrt{u}}\left(2 \sec ^2 x\right)\)
\(=\frac{\sec ^2 x}{\sqrt{1+2 \tan x}}\)

\(y=x^2+{1\over x^2}-2=x^2+x^{-2}-2\)
\(\frac{d y}{d x}=2 x-2 x^{-2-1}=2 x-\frac{2}{x^3}\)

\({dy\over dx}=-4\ cosec \ x.cot \ x -{1\over x}-2e^x\)

\(\frac{d y}{d x}=e^x+\cos x\)

\(f(x)= \begin{cases}x, & x \leq 1 \\ x^2, & x>1\end{cases}\)
\(f^{\prime}\left(1^{-}\right)=\lim _{x \rightarrow 1^{-}} \frac{f(\dot{x})-f(1)}{x-1}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{(x-1)}{(x-1)}=\lim _{x \rightarrow 1^{-}}(1)=1\)

\(f^{\prime}\left(1^{+}\right)=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{x^2-1}{x-1}\) \(\begin{aligned} &{\left[\because f(x)=x^2\right.} &\left.f(1)=1^2=1\right] \end{aligned}\)
\(=\lim _{x \rightarrow 1^{+}} \frac{(x+1)(x-1)}{(x-1)}=1+1=2\)
\(f^{\prime}\left(1^{+}\right)=2\)
\(\therefore f^{\prime}\left(1^{-}\right) \neq f^{\prime}\left(1^{+}\right)(\because \text { by }(1) \&(2))\)
\(\therefore\) It is not differentiable.

\(f(x)=\sqrt{1-x^2}\)
\(f^{\prime}\left(1^{-}\right)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{1-x^2}-0}{x-1} \quad \begin{aligned} f(x) &=\sqrt{1-x^2} \\ f(1) &=\sqrt{1-1} \\ &=0 \end{aligned}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{1-x^2}}{x-1}=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{(1-x)(1+x)}}{x-1}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{1-x} \sqrt{1+x}}{x-1}\)
\(\therefore f^{\prime}(x)=\lim _{x \rightarrow 1^{-}} \frac{+\sqrt{1-x} \sqrt{1+x}}{-(1-x)}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{1-x} \sqrt{1+x}}{-\sqrt{1-x} \sqrt{1-x}}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{1+x}}{-\sqrt{1-x}} \rightarrow-\infty\)
\(\therefore f^{\prime}(x) \rightarrow-\infty^{-} \text {as } x \rightarrow 1^{-}\)
It is not diferentiable.

\(f(x)=|x-1|= \begin{cases}-(x-1) & \text { if } x<1 \\ (x-1) & \text { if } x>1\end{cases}\)

We know that this function is continuous at x = 1
But \(f^{\prime}\left(1^{-}\right)=\lim _{z \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{-(x-1)-[-(1-1)]}{x-1}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{-x+1-0}{x-1}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{-(x-1)}{(x-1)}=\lim _{x \rightarrow 1^{-}}(-1)\)
\(f^{\prime}\left(1^{-}\right)=-1\ ......(1)\)
\(f^{\prime}\left(1^{+}\right)=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\)
\(=\lim _{x \rightarrow 1^{+}} \frac{(x-1)-0}{x-1}\)
\(=\lim _{x \rightarrow 1^{+}} \frac{(x-1)}{(x-1)}=\lim _{x \rightarrow 1^{+}}(1)\)
\(f^{\prime}\left(1^{+}\right)=1\ ......(2)\)
\(f^{\prime}\left(1^{-}\right) \neq f^{\prime}\left(1^{+}\right) \quad(\because \text { by }(1) \&(2))\)
\(\therefore f\) is not differentiable at x = 1.

\(\int \frac{\sec x}{\sqrt{\cos 2 x}} d x =\int \frac{\sec x}{\sqrt{\cos ^{2} x-\sin ^{2} x}} d x \)
\(=\int \frac{\sec x}{\sqrt{\cos ^{2} x\left(1-\frac{\left.\sin ^{2} x\right)}{\cos ^{2} x}\right)} d x} \)
\(=\int \frac{\sec x}{\cos x \sqrt{1-\tan ^{2} x}} d x \)
\(=\int \frac{1}{\sqrt{1-\tan ^{2} x}} \times \sec ^{2} x d x \)
\(=\int \frac{1}{\sqrt{1-u^{2}}} \times d u, u=\tan x \)
\(=\sin ^{-1} u+c \)
\(=\sin ^{-1}(\tan x)+c \)

\(\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x =\int \frac{x^{6}-x^{5}}{x^{4}-x^{3}} d x \)
\(=\int \frac{x^{2}\left(x^{4}-x^{3}\right)}{x^{4}-x^{3}} d x \)
\(=\int x^{2} d x=\frac{x^{3}}{3}+c \)

\(\sin 3 x =3 \sin x-4 \sin ^{3} x \)
\(\therefore 4 \sin ^{3} x =3 \sin x-\sin 3 x \)
\(\therefore \sin ^{3} x =\frac{1}{4}[3 \sin x-\sin 3 x] \)
\(\therefore \int \sin ^{3} x d x =\frac{1}{4}\left[-3 \cos x+\frac{\cos 3 x}{3}\right]+c \)
\(=\frac{-3}{4} \cos x+\frac{\cos 3 x}{12}+c \)

\(f(x)=|x-1|+|x-3|+\sin x\)
\(\text { Since } \sin x \text { is differentiable everywhere }\)

At x = 1 and x = 3. The graph admit cups.
\(\therefore\) The derivative is not exist.
\(\therefore\) The number of points in R is 2.

Given f is differentiable
\(\therefore f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right)=1 \)
\(f^{\prime}\left(1^{-}\right) =\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x+1}=\lim _{x \rightarrow 1^{-}} \frac{\left(a x^{2}-b\right)-(a-b)}{x+1} \)
\(=\lim _{x \rightarrow 1^{-}} \frac{a x^{2}-b-a+b}{x+1}=\lim _{x \rightarrow 1^{-}} \frac{a\left(x^{2}-1\right)}{x+1} \)
\(=\lim _{x \rightarrow 1^{-}}-a(x+1) \)
\(=a(1+1)=2 a \)
\(\therefore f^{\prime}\left(1^{+}\right) =\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{\frac{1}{x}-1}{x-1} \)
\(=\lim _{x \rightarrow 1^{+}} \frac{1-x}{x(x-1)}=\lim _{x \rightarrow 1^{+}} \frac{-1}{x}=-1\)
\(\therefore 2 a =-1 \)
\(a =\frac{-1}{2} \)
\(\text { and } f(1)=1\)
\(a-b=1 \)
\(-1 / 2-1=b \)
\(b=-3 / 2 \)

\(\text { Given } f(x)=x+2\)
\(f^{\prime}(x) =1 \)
\(f^{\prime}(f(x)) =f^{\prime}(x+2)=1 \)

\( y=\log _{10} x=\frac{1}{\log _x 10}\)
Diff  w. r. to \(\log _x 10\)
\(y^{\prime}=\frac{-1}{\left(\log _x 10\right)^2}=-\left(\log _{10} x\right)^2 \)

\(x =a \sin \theta, y=b \cos \theta \)
\(\frac{d x}{d \theta} =a \cos \theta, \frac{d y}{d \theta}=-b \sin \theta\)
\(\frac{d y}{d x} =\frac{d y / d \theta}{d x / d \theta}=\frac{-b \sin \theta}{a \cos \theta}=\frac{-b}{a} \tan \theta \)
\(\frac{d^{2} y}{d x^{2}} =\frac{-b}{a} \sec ^{2} \theta \frac{d \theta}{d x}\)
\(=\frac{-b}{a} \sec ^{2} \theta \cdot \frac{1}{a \cos \theta} \)
\(= \frac{-b}{a^{2}} \sec ^{3} \theta \)

\(y=f\left(x^{2}+2\right) \)
\(\frac{d y}{d x} =f^{\prime}\left(x^{2}+2\right)(2 x) \)
\(\text { At } x =1, \frac{d y}{d x}=f^{\prime}(1+2)(2)=f^{\prime}(3)(2) \)
\(=5(2)=10 \)

\(\frac{d}{d x}\left(\frac{2}{\pi} \sin x\left(\frac{\pi^{\circ}}{180}\right)\right) =\frac{2}{\pi} \cos \frac{\pi x^{\circ}}{180^{\circ}} \frac{\pi}{180^{\circ}} \)
\(=\frac{1}{90} \cos x^{\circ} \)